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Question Number 187827 by Rupesh123 last updated on 22/Feb/23

	Answered by cortano12 last updated on 23/Feb/23
	$(1)tan 30°=(4/a)⇒a=4(√3) (2) tan 22,5^o =(4/b) ; tan 45°=((2tan 22,5^o )/(1−tan^2 22,5^o )) ⇒tan^2 22,5^o +2tan 22,5^o −1=0 ⇒tan 22,5^o =((−2+2(√2))/2)=(√2)−1 ⇒b=(4/( (√2)−1))=4(√2)+4 (3) tan 37,5^o =(4/c) ; tan 75^o =((2tan 37,5^o )/(1−tan^2 37,5^o )) ⇒2+(√3)=((2tan 37,5^o )/(1−tan^2 37,5^o )) ⇒(2+(√3))tan^2 37,5^o +2tan 37,5^o −(2+(√3))=0 ⇒tan 37,5^o =((−2+(√(11+2(√(12)))))/(2(2+(√3)))) ⇒tan 37,5^o =((5(√3)−8)/2) ⇒c=(8/( 5(√3)−2))=(8/(71))(5(√3)+2) ∴ sides length of the triangle = { ((a+b=4(√3)+4(√2)+4)),((a+c=4(√3)+((16+40(√3))/(71))=((324(√3)+16)/(71)))),((b+c=4(√2)+4+((40(√3)+16)/(71))=((300+284(√2)+40(√3))/(71)))) :}$
	$(1) \tan 30 ° = \frac{4}{a} \Rightarrow a = 4 \sqrt{3}$ $(2) \tan 22, 5^{o} = \frac{4}{b}; \tan 45 ° = \frac{2 \tan 22, 5^{o}}{1 - \tan^{2} 22, 5^{o}}$ $\Rightarrow \tan^{2} 22, 5^{o} + 2 \tan 22, 5^{o} - 1 = 0$ $\Rightarrow \tan 22, 5^{o} = \frac{- 2 + 2 \sqrt{2}}{2} = \sqrt{2} - 1$ $\Rightarrow b = \frac{4}{\sqrt{2} - 1} = 4 \sqrt{2} + 4$ $(3) \tan 37, 5^{o} = \frac{4}{c}; \tan 75^{o} = \frac{2 \tan 37, 5^{o}}{1 - \tan^{2} 37, 5^{o}}$ $\Rightarrow 2 + \sqrt{3} = \frac{2 \tan 37, 5^{o}}{1 - \tan^{2} 37, 5^{o}}$ $\Rightarrow (2 + \sqrt{3}) \tan^{2} 37, 5^{o} + 2 \tan 37, 5^{o} - (2 + \sqrt{3}) = 0$ $\Rightarrow \tan 37, 5^{o} = \frac{- 2 + \sqrt{11 + 2 \sqrt{12}}}{2 (2 + \sqrt{3})}$ $\Rightarrow \tan 37, 5^{o} = \frac{5 \sqrt{3} - 8}{2}$ $\Rightarrow c = \frac{8}{5 \sqrt{3} - 2} = \frac{8}{71} (5 \sqrt{3} + 2)$ $∴ sides length of the triangle$ $= {\begin{cases} a + b = 4 \sqrt{3} + 4 \sqrt{2} + 4 \\ a + c = 4 \sqrt{3} + \frac{16 + 40 \sqrt{3}}{71} = \frac{324 \sqrt{3} + 16}{71} \\ b + c = 4 \sqrt{2} + 4 + \frac{40 \sqrt{3} + 16}{71} = \frac{300 + 284 \sqrt{2} + 40 \sqrt{3}}{71} \end{cases}$
	Commented by Rupesh123 last updated on 22/Feb/23
	Very nice work!
	Answered by nikif99 last updated on 22/Feb/23

	$L e t i n c e n t e r : I . L e t I D r a d i u s = 4 c m$ $∡ C = 75 °$ $A D = \frac{I D}{\tan 22.5} = 4 (\sqrt{2} + 1)$ $B D = \frac{I D}{\tan 30} = 4 \sqrt{3}$ $A B = 4 (1 + \sqrt{2} + \sqrt{3}) c m (1)$ $\frac{A C}{\sin 60} = \frac{A B}{\sin 75} = \frac{B C}{\sin 45} (2)$ $(1) (2) \Rightarrow A C = 4 (3 - \sqrt{3} + \sqrt{6}) c m$ $B C = 4 (2 - \sqrt{2} + \sqrt{6}) c m$
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